Equations of Straight Lines

Compare with y=mx+c: the gradient is the number multiplying x (with its sign), wherever it is written.

1. y=4-\frac{1}{2}x is the same as y=-\frac{1}{2}x+4
2. So m=-\frac{1}{2} and c=4
3. The line crosses the y-axis where x=0, at (0,\,4)
4. Answer: the gradient is -\frac{1}{2} and the line crosses the y-axis at (0,\,4)

Put x=0 into the equation to find where the line crosses the y-axis, then put y=0 to find where it crosses the x-axis.

1. When x=0: -3y=12, so y=-4; the line crosses the y-axis at (0,\,-4)
2. When y=0: 4x=12, so x=3; the line crosses the x-axis at (3,\,0)
3. Check the gradient: 3y=4x-12, so y=\frac{4}{3}x-4, a positive gradient
4. Answer: C

Gradient =\dfrac{\text{change in }y}{\text{change in }x}; take care subtracting the negative coordinates.

1. Change in y: 3-(-7)=10
2. Change in x: 3-(-1)=4
3. Gradient =\dfrac{10}{4}
4. Answer: \dfrac{5}{2}

Rearrange the equation of L into the form y=mx+c first; perpendicular gradients multiply to -1.

1. 3y=-4x+7
2. y=-\frac{4}{3}x+\frac{7}{3}, so the gradient of L is -\frac{4}{3}
3. Perpendicular gradient: change the sign and turn the fraction upside down
4. Check: -\frac{4}{3}\times\frac{3}{4}=-1
5. Answer: \dfrac{3}{4}

Substitute x=10 and y=3 into y=-\frac{2}{5}x+c to find c.

1. y=-\frac{2}{5}x+c
2. At (10,\,3): 3=-\frac{2}{5}\times10+c
3. 3=-4+c, so c=7
4. Answer: y=-\frac{2}{5}x+7

Rearrange each equation into the form y=mx+c and compare the gradients; parallel lines have equal gradients.

1. 2y=5x-3 gives y=\frac{5}{2}x-\frac{3}{2}, so the gradient is \frac{5}{2}
2. 5y=2x+1: gradient \frac{2}{5} (upside down)
3. 10x-4y=7: 4y=10x-7, so y=\frac{5}{2}x-\frac{7}{4}, gradient \frac{5}{2}
4. 5x+2y=6: gradient -\frac{5}{2} (wrong sign)
5. y-5x=2: gradient 5
6. Answer: 10x-4y=7

Multiply both sides by 3 to clear the fraction, then collect the x and y terms on one side.

1. Multiply both sides by 3: 3y-9=-2(x+6)
2. 3y-9=-2x-12
3. Add 2x and 9 to both sides: 2x+3y=-3
4. Answer: 2x+3y=-3

Rearrange ky-10x=3 into the form y=mx+c; parallel lines have the same gradient.

1. Rearrange: ky=10x+3, so y=\dfrac{10}{k}x+\dfrac{3}{k}
2. The gradient is \dfrac{10}{k}
3. Parallel lines have equal gradients: \dfrac{10}{k}=4
4. k=\dfrac{10}{4}=\dfrac52
5. Answer: k=\dfrac52

Write the equation using y-y_1=m(x-x_1), then put y=0 because P is on the x-axis.

1. y-(-2)=\frac{3}{5}(x-4), so y+2=\frac{3}{5}(x-4)
2. On the x-axis y=0: 2=\frac{3}{5}(x-4)
3. Multiply by \frac{5}{3}: x-4=\frac{10}{3}
4. x=4+\frac{10}{3}=\frac{22}{3}
5. Answer: P=\left(\frac{22}{3},\,0\right)

Find the gradient of x+4y=10, then the gradient AB must have; write the gradient of AB in terms of k and set them equal.

1. x+4y=10 gives y=-\frac{1}{4}x+\frac{5}{2}, gradient -\frac{1}{4}
2. So the gradient of AB is 4 (because -\frac{1}{4}\times4=-1)
3. Gradient of AB=\dfrac{-7-9}{3-k}=\dfrac{-16}{3-k}
4. \dfrac{-16}{3-k}=4, so -16=12-4k
5. 4k=28
6. Answer: k=7

Write the equation of line A in the form y=mx+c, then substitute it into the equation of line B.

1. Line A: y=3x-5
2. Substitute into 4x+5y=13: 4x+5(3x-5)=13
3. 4x+15x-25=13
4. 19x=38, so x=2
5. y=3\times2-5=1
6. Answer: (2,\,1)

Find the gradient first (change in y divided by change in x), then use y-y_1=m(x-x_1) and clear the fraction.

1. Gradient =\dfrac{8-(-1)}{-4-8}=\dfrac{9}{-12}=-\dfrac34
2. Use y-y_1=m(x-x_1) with (8,\,-1): y+1=-\dfrac34(x-8)
3. Multiply both sides by 4: 4y+4=-3(x-8)=-3x+24
4. Collect x and y on the left: 3x+4y=20
5. Check with (-4,\,8): -12+32=20 ✓
6. Answer: 3x+4y=20

Substitute x=5 and y=4 to find p first; the line crosses the y-axis where x=0.

1. Substitute (5,\,4): 5p-8=3p
2. 2p=8, so p=4
3. The line is 4x-2y=12
4. On the y-axis x=0: -2y=12, so y=-6
5. Answer: (0,\,-6)

Rearrange 5x+2y=3 to find the gradient of L; the gradient of M is its negative reciprocal.

1. 2y=-5x+3, so y=-\frac{5}{2}x+\frac{3}{2}; the gradient of L is -\frac{5}{2}
2. The gradient of M is \frac{2}{5} (because -\frac{5}{2}\times\frac{2}{5}=-1)
3. y=\frac{2}{5}x+c through (-5,\,4): 4=\frac{2}{5}\times(-5)+c
4. 4=-2+c, so c=6
5. Answer: y=\frac{2}{5}x+6

DC is parallel to AB, so it has the same gradient as AB and passes through D.

1. Gradient of AB=\dfrac{-3-1}{7-1}=\dfrac{-4}{6}=-\dfrac{2}{3}
2. DC is parallel to AB, so its gradient is also -\frac{2}{3}
3. y=-\frac{2}{3}x+c through D(-3,\,5): 5=-\frac{2}{3}\times(-3)+c
4. 5=2+c, so c=3
5. Answer: y=-\frac{2}{3}x+3

Perpendicular gradients multiply to -1; write the gradient of L_2 as a fraction and rationalise its denominator. Answer part (a) before attempting part (b).

Part (a)

1. Gradient of L_2=-\dfrac{1}{2+\sqrt{3}}
2. Multiply top and bottom by 2-\sqrt{3}
3. Bottom: (2+\sqrt{3})(2-\sqrt{3})=4-3=1
4. So the gradient is -(2-\sqrt{3})
5. Answer: -2+\sqrt{3}

Part (b)

1. L_2: y=(\sqrt{3}-2)x+4
2. On the x-axis y=0: (\sqrt{3}-2)x=-4
3. x=\dfrac{-4}{\sqrt{3}-2}=\dfrac{4}{2-\sqrt{3}}
4. Multiply top and bottom by 2+\sqrt{3}: the bottom becomes 4-3=1
5. x=4(2+\sqrt{3})
6. Answer: 8+4\sqrt{3}

Write the gradients of PQ and QR in terms of a, then use the fact that the product of perpendicular gradients is -1.

1. Gradient of PQ=\dfrac{4-0}{2-a}=\dfrac{4}{2-a}
2. Gradient of QR=\dfrac{8-4}{a-2}=\dfrac{4}{a-2}
3. Perpendicular: \dfrac{4}{2-a}\times\dfrac{4}{a-2}=-1
4. (2-a)(a-2)=-(a-2)^2, so \dfrac{16}{-(a-2)^2}=-1
5. (a-2)^2=16, so a-2=4 or a-2=-4
6. Answer: a=6 or a=-2

A tangent to a circle is perpendicular to the radius at that point, so start with the gradient of OP. Answer part (a) before attempting part (b).

Part (a)

1. Gradient of the radius OP=\dfrac{4}{2}=2
2. The tangent is perpendicular to the radius, so its gradient is -\frac{1}{2}
3. y-4=-\frac{1}{2}(x-2)
4. Multiply by 2: 2y-8=-x+2
5. Answer: x+2y=10

Part (b)

1. From 2x-y=10: y=2x-10
2. Substitute into x+2y=10: x+2(2x-10)=10
3. 5x-20=10, so x=6
4. y=2\times6-10=2
5. Answer: R=(6,\,2)

Perpendicular gradients multiply to -1; for the area, find where L_1 crosses the x-axis and use the x-axis as the base. Answer part (a) before attempting part (b).

Part (a)

1. The gradient of L_1 is 2, so the gradient of L_2 is -\frac{1}{2}
2. y=-\frac{1}{2}x+c through (12,\,0): 0=-6+c
3. c=6
4. Answer: y=-\frac{1}{2}x+6

Part (b)

1. At P: 2x-4=-\frac{1}{2}x+6
2. \frac{5}{2}x=10, so x=4
3. y=2\times4-4=4
4. Answer: P=(4,\,4)

Part (c)

1. L_1 crosses the x-axis where 2x-4=0, at (2,\,0)
2. The base of the triangle runs from (2,\,0) to (12,\,0): length 10
3. The height is the y-coordinate of P: 4
4. Area =\frac{1}{2}\times10\times4
5. Answer: 20

R is \dfrac45 of the way from P to Q; the gradient of L is the negative reciprocal of the gradient of PQ.

Part (a)

1. From P to Q: x goes up by 6-(-4)=10 and y changes by -2-3=-5
2. PR:RQ=4:1, so R is \dfrac45 of the way from P to Q
3. \dfrac45\times10=8 and \dfrac45\times(-5)=-4
4. R=(-4+8,\ 3-4)
5. Answer: (4,\,-1)

Part (b)

1. Gradient of PQ=\dfrac{-2-3}{6-(-4)}=\dfrac{-5}{10}=-\dfrac12
2. Perpendicular gradient =2 (because -\dfrac12\times2=-1)
3. Through R(4,\,-1): y+1=2(x-4)
4. y=2x-8-1
5. Answer: y=2x-9

Part (c)

1. Substitute y=2x-9 into x+3y=8: x+3(2x-9)=8
2. x+6x-27=8
3. 7x=35, so x=5
4. y=2\times5-9=1
5. Answer: (5,\,1)